JEE Main 2016 — Parabola Question with Solution

$y^2=8x$ is the equation of the given parabola. If $P$ is a point at a minimum distance from $\text{'}\left(0,-6\right)\text{'}$, then it should be normal to the parabola at $P$.

Normal to parabola $y^2=8x$ is 

$y=mx-2·2·m-2·m^3$

It passes through $\left(0,-6\right)$

$\Rightarrow m^3+2m-3=0\Rightarrow m=1$

$P\left(a{m}^2,-2am\right)=P(2,-4)$

Equation of circle with centre $P$ and passes through $\text{'}C\text{'}$ is

${\left(x-2\right)}^2+{\left(y+4\right)}^2=8$

$\Rightarrow x^2+y^2-4x+8y+12=0$